The question I am asking is to solve the equation $x^4-4x-1=0$, I need an exact answer. What I have done was found out that it equals $(x^2+1)^2 - 2(x+1)^2 =0$. Anybody help me, please?
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1Hint: you now have an expression of the form x^2 - y^2 = 0. How does that simplify? – Julia Hayward Jul 21 '20 at 14:45
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1If you edit your post and surround your math with dollar signs, like this: $x^4-4x-1=0$, it will better fit site standards for writing math. – David Diaz Jul 21 '20 at 14:47
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@JuliaHayward So basically it's a difference of squares now. – Jul 21 '20 at 14:49
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Yes, what can we say about the relation between $x$ and $y$? – Stockfish Jul 21 '20 at 14:49
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So you know
$$\begin{align} x^4-4x-1&=(x^2+1)^2-2(x+1)^2\\ &=(x^2+1)^2-[\sqrt2(x+1)]^2. \end{align}$$
Factorise this difference of two squares to get two quadratic factors, hence the four roots.
Shaun
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user10354138
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